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The following problem has kept me pondering for a while now and since I can't get through, I'm posting it here.

Say that you can draw a number $x_i$ uniformly from the set $\mathcal{X} \sim \{0,0,1,2,3,4,5,6\}$. The sum of those numbers we call $X$. We stop drawing numbers when $X$ is at least $5$, the amount of numbers drawn we will call $N$. I would write the stopping condition as:

$$X = \sum_{i=0}^{N}x_i \geq 5 $$

What is the expected value of numbers we need to draw, $E[N]$?

I'll attempt some visualisations based on empirical outcomes.

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Define the function $h(x)=\mathbb{E}_x(N)$ to be the expected number of steps needed, if we start with a sum of $x$. First step analysis gives

\begin{eqnarray*} h(4)&=&1+h(4)/4 \\ h(3)&=&1+h(3)/4+h(4)/8\\ h(2)&=&1+h(2)/4+h(3)/8+h(4)/8\\ h(1)&=&1+h(1)/4+h(2)/8+h(3)/8+h(4)/8\\ h(0)&=&1+h(0)/4+h(1)/8+h(2)/8+h(3)/8+h(4)/8. \end{eqnarray*}

Solving this system we get $E(N)=h(0)={2401\over 972}\approx 2.47.$

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It is possible to solve a more general problem. We are looking for $$N(t)=\min \left\{ n : \sum_{i=1}^n x_i >t\right\}.$$ Now set $$m(t)=\mathbb{E}[N(t)]$$ We can condition on $x_1$ to obtain $$m(t)=\cases{ 1 & if $\quad x_1\ge t$\cr m(t-x_1) & if $\quad x_1

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    This looks like a wonderful solution and I have been trying to understand it but I am having some difficulties with it. Could you explain why you can condition $x_i$ to define $m(t)$ like that? – 2017-02-13